- Gas within a chamber passes through a cycle ABCA. • Process AB is isometric pr
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Question
- Gas within a chamber passes through a cycle ABCA.
• Process AB is isometric process. The heat absorbed during process AB is 50.0J.
• Process BC is adiabatic process. The work done by the gas during process BC is 15.0 J.
• Process CA is isobaric process with pressure of 1000 N/m2. The volume decreases 0.02 m3 during process CA.
What is the energy transfer as heat during process CA?
- An ideal gas in a rigid container is initially at a temperature of 27oC and a pressure 1.0 x 105 Pascal. When the gas is heated to a temperature of 77oC, what is the pressure of the gas?
Explanation / Answer
from the given data,
WAB = 50 J, dUAB = 0, QAB = 50 J
WBC = 15 J, QBC = 0, dUBC = -15 J
WCA = P*(V2-V1)
= -1000*0.02
= -20 J
in a complete cycle, change in internal energy, dU = 0
dUAB + dUBC + dUCA = 0
dUCA = -dUAB - dUBC
= 0 - (-15)
= 15 J
use first law of thermodynamics,
QCA = WCA + dUCA
= -20 + 15
= -5 J <<<<<<<<<<<<--------------Answer
here negative sign indicate system loses heat energy.
2) T1 = 27 C = 27 + 273 = 300 K
T2 = 77 C + 77 + 273 = 350 K
P1 = 1*10^5 pa
P2 = ?
at constant volume, P/T constant
so,
P2/T2 = P1/T1
P2 = P1*(T2/T1)
= 1*10^5*(350/300)
= 1.167*10^5 pa <<<<<<<<<<<--------------Answer
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